Through a point O$(0,0)$ , a line L is drawn which meets two other lines $3x+4y=5$ and $x+2y=3$ at points P and Q. Find the locus of N on the variable line if ON is the arithmetic mean of OP AND OQ.
A try
I used the parametric form of an arbitrary line and calculated the mean distance to be $$\frac{0.5(19 \cos x+25 \sin x)}{(3\cos ^2 x+10 \sin^2 x+ 11 \sin 2x)}$$ I was unable to use this to write the locus . I was looking for a hint or an alternate approach to the problem.
On
Say $y=lx$ is variable line through $(0,0)$. Then $$P\left({5\over 4l+3}, {5l\over 4l+3}\right)$$ and $$Q\left(\frac{3}{2l+1},\frac{3l}{2l+1}\right)$$ also $y_N =lx_N$ and $$2x_N = {5\over 4l+3} + {3\over 2l+1}={22l+14\over 8l^2+10l+3} $$
Rewrite $x \to x_N$ and $y\to y_N$. Then we have: $$2x = {22lx^2+14x^2\over 8l^2x^2+10lx^2+3x^2} = {22yx+14x^2\over 8y^2+10yx+3x^2}$$ and we get the hyperbola:$$8y^2+10xy+3x^2 = 11y+7x$$
On
For any line $\ell : ax + by = c$ in the plane, $n = (a,b)$ is a normal vector for that line. In terms of $n$, we can rewrite the condition that a point $p = (x,y)$ lying on line $\ell$ as $n \cdot p = c$.
Let's say we have two lines $\ell_1, \ell_2$ which are disjoint from $O = (0,0)$. If we draw a line $\ell_3$ through $O$ along some direction $d$ and let it hit line $\ell_1$ at $p_1 = \lambda_1 d$ and hit the line $\ell_2$ at $p_2 = \lambda_2 d$. We will have $$ \begin{cases} n_1 \cdot p_1 = \lambda_1 n_1 \cdot d = c_1\\ n_2 \cdot p_2 = \lambda_2 n_2 \cdot d = c_2 \end{cases} \quad\implies\quad \begin{cases} \lambda_1 = \frac{c_1}{n_1\cdot d}\\ \lambda_2 = \frac{c_2}{n_2\cdot d} \end{cases} $$ Let $p_3$ be the mid-point of $p_1, p_2$, we have $$p_3 = \frac12(p_1+p_2) = \frac12(\lambda_1+\lambda_2) d = \frac12\left(\frac{c_1}{n_1\cdot d}+ \frac{c_2}{n_2\cdot d}\right) d = \frac12\left(\frac{c_1}{n_1\cdot p_3}+ \frac{c_2}{n_2\cdot p_3}\right)p_3$$ This leads to
$$\frac12\left(\frac{c_1}{n_1\cdot p_3}+ \frac{c_2}{n_2\cdot p_3}\right) = 1 \quad\iff\quad \left(n_1 \cdot p_3 - \frac{c_1}{2}\right)\left(n_2 \cdot p_3 - \frac{c_2}{2}\right) = \frac{c_1c_2}{4}$$
For the problem at hand, we can set $\ell_1$ to the line $3x + 4y = 5$ and $\ell_2$ to the line $x+2y = 3$. i.e. $$n_1 = (3,4), c_1 = 5\quad\text{ and }\quad n_2 = (1,2), c_2 = 3$$ $P, Q, N$ will take the roles of $p_1, p_2, p_3$ in above discussion. As a result, $N = (x,y)$ satisfy: $$ \left(3x+4y - \frac{5}{2}\right)\left(x + 2y - \frac{3}{2}\right) = \frac{15}{2} \quad\iff\quad 3x^2 + 10xy + 8y^2 -7x-11y = 0 $$ This is the equation of a hyperbola having lines $\tilde{\ell}_1 : 3x+4y = \frac{5}{2}$ and $\tilde{\ell}_2 : x + 2y - \frac{3}{2}$ as asymptotes. Geometrically, $\tilde{\ell}_1$ is a line parallel to $\ell_1$ and mid way between $O$ and $\ell_1$. Same thing happens to asymptote $\tilde{\ell}_2$.
One branch of this hyperbola passes through $O$. The other branch passes through $X$, the intersection of $\ell_1$ and $\ell_2$. The pair of lines $\ell_1$ and $\ell_2$ divides the plane into $4$ components. When $O$ lies between $P, Q$, $N$ falls on the branch passes through $O$. This branch is contained in the interior of that component containing $O$. Otherwise, $N$ falls on the branch passes through $X$. With the exception of $X$, this branch is contained in the union of interior of the two components adjacent to the one containing $O$.
for both lines we get $$y=-\frac{3}{4}x+\frac{5}{4}$$ and $$y=-\frac{1}{2}x+\frac{3}{2}$$ for l we take $$y=mx$$ and we get the cordinates of both points $P,Q$ as $$P\left(\frac{5}{3+4m};\frac{5m}{3+4m}\right)$$ $$Q\left(\frac{3}{1+2m};\frac{3m}{1+2m}\right)$$ Can you finish? then $N$ has the coordinates $$N\left(\frac{x_P+x_Q}{2};\frac{y_Q+y_P}{2}\right)$$ setting $$x=\left(\frac{5}{3+4m}+\frac{3}{1+2m}\right)\cdot \frac{1}{2}$$ after some algebra we obtain $$m=\frac{9-10x-\sqrt{81+12x+4x^2}}{16x}$$ or $$m=\frac{9-10x+\sqrt{81+12x+4x^2}}{16x}$$ can you now finish?